Inspired By Comrade Otto
f ( 0 ) f ( 1 ) f ( 2 ) f ( 3 ) f ( 4 ) f ( 5 ) f ( 6 ) = = = = = = = 2 0 2 1 2 2 2 3 2 4 2 5 2 6
If f ( x ) is a degree 6 polynomial satisfying the above values, then what is the value of f ( 1 3 ) ?
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2 solutions
Can someone help me f(3)-1=7 but shouldn't it be divisible by 3 as the constant term of polynomial is 1
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You are making the assumption that the coefficients of the polynomial are integers, which they need not be. E.g. For f ( x ) = 3 1 x + 1 , clearly f ( 3 ) − f ( 0 ) is not divisible by 3.
Would u please elaborate , that how did u reach the f(x)
Plz explain it
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Explain what? Which step?
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If u plz dont mind. Can u please explain how we write the 1st step .
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@Pranav Bansal – See the examples . It is an identity of binomial theorem.
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@Aditya Kumar – It is in class 11th ?
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@Chirayu Bhardwaj – Yes. It is a basic identity.
How do you get the second summation
We use the method of differences .
Since p(x) is 6th degree, the sixth differences D^6 p(x) must be the same number, namely 1 (see below).
1 2 4 8 16 32 64
1 2 4 8 16 32
1 2 4 8 16
1 2 4 8
1 2 4
1 2
1
By calculus of finite differences, the polynomial is p(13) = (I+D)^13 p(0) = p(0)+C(13, 1) Dp(0)+C(13, 2) D^2p(0)+...+C(13, 12)*D^12p(0)+D^13p(0) = 1+C(13, 1)+C(13, 2)+C(13, 3)+...+C(13, 6) = (1+1)^13/2 = 4096.
In general, p(x) = 1+C(x, 1)+C(x, 2)+...+C(x, 6) = 1+x+x(x-1)/2+x(x-1)(x-2)/6+x(x-1)(x-2)(x-3)/24+x(x-1)(x-2)(x-3)(x-4)/120+x(x-1)(x-2)(x-3)(x-4)(x-5)/720.
So what is the polynomial.? Obviously, the first term is 1. I suppose the above would lead to a system of 6 linear equations involving the 6 coefficients which could be solved by matrix algebra
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The point is that we don't need to find the exact polynomial (if we're only interested in the next value). Check out the wiki page method of differences and learn from it!
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We define the polynomial to be a x 6 + b x 5 + c x 4 + d x 3 + e x 2 + f x + g Through plugging in the values 0 through 6, we have a system: ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ a 0 6 + b 0 5 + c 0 4 + d 0 3 + e 0 2 + f 0 + g a 1 6 + b 1 5 + c 1 4 + d 1 3 + e 1 2 + f 1 + g a 2 6 + b 2 5 + c 2 4 + d 2 3 + e 2 2 + f 2 + g a 3 6 + b 3 5 + c 3 4 + d 3 3 + e 3 2 + f 3 + g a 4 6 + b 4 5 + c 4 4 + d 4 3 + e 4 2 + f 4 + g a 5 6 + b 5 5 + c 5 4 + d 5 3 + e 5 2 + f 5 + g a 6 6 + b 6 5 + c 6 4 + d 6 3 + e 6 2 + f 6 + g = 1 = 2 = 4 = 8 = 1 6 = 3 2 = 6 4
Rewriting yields:
⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ 0 6 a + 0 5 b + 0 4 c + 0 3 d + 0 2 e + 0 f + g 1 6 a + 1 5 b + 1 4 c + 1 3 d + 1 2 e + 1 f + g 2 6 a + 2 5 b + 2 4 c + 2 3 d + 2 2 e + 2 f + g 3 6 a + 3 5 b + 3 4 c + 3 3 d + 3 2 e + 3 f + g 4 6 a + 4 5 b + 4 4 c + 4 3 d + 4 2 e + 4 f + g 5 6 a + 5 5 b + 5 4 c + 5 3 d + 5 2 e + 5 f + g 6 6 a + 6 5 b + 6 4 c + 6 3 d + 6 2 e + 6 f + g = 1 = 2 = 4 = 8 = 1 6 = 3 2 = 6 4 Simplifying and placing the values and putting them in a matrix
⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 0 1 6 4 7 2 9 4 0 9 6 1 5 6 2 5 4 6 6 5 6 0 1 3 2 2 4 3 1 0 2 4 3 1 2 5 7 7 7 6 0 1 1 6 8 1 2 5 6 6 2 5 1 2 9 6 0 1 8 2 7 6 4 1 2 5 2 1 6 0 1 4 9 1 6 2 5 3 6 0 1 2 3 4 5 6 1 1 1 1 1 1 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ a b c d e f g ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 1 2 4 8 1 6 3 2 6 4 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤
⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ a b c d e f g ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 0 1 6 4 7 2 9 4 0 9 6 1 5 6 2 5 4 6 6 5 6 0 1 3 2 2 4 3 1 0 2 4 3 1 2 5 7 7 7 6 0 1 1 6 8 1 2 5 6 6 2 5 1 2 9 6 0 1 8 2 7 6 4 1 2 5 2 1 6 0 1 4 9 1 6 2 5 3 6 0 1 2 3 4 5 6 1 1 1 1 1 1 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ − 1 ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 1 2 4 8 1 6 3 2 6 4 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤
⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ a b c d e f g ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 0 . 0 0 1 3 8 9 − 0 . 0 2 9 1 7 0 . 2 4 3 0 5 6 − 1 . 0 2 0 8 3 2 . 2 5 5 5 5 6 − 2 . 4 5 1 − 0 . 0 0 8 3 3 0 . 1 6 6 6 6 7 − 1 . 2 9 1 6 7 4 . 8 3 3 3 3 3 − 8 . 7 6 0 0 . 0 2 0 8 3 3 − 0 . 3 9 5 8 3 2 . 8 5 4 1 6 7 − 9 . 6 0 4 1 7 1 4 . 6 2 5 − 7 . 5 0 − 0 . 0 2 7 7 8 0 . 5 − 3 . 3 6 1 1 1 1 0 . 3 3 3 3 3 − 1 4 . 1 1 1 1 6 . 6 6 6 6 6 7 0 0 . 0 2 0 8 3 3 − 0 . 3 5 4 1 7 2 . 2 2 9 1 6 7 − 6 . 3 9 5 8 3 8 . 2 5 − 3 . 7 5 0 − 0 . 0 0 8 3 3 0 . 1 3 3 3 3 3 − 0 . 7 9 1 6 7 2 . 1 6 6 6 6 7 − 2 . 7 1 . 2 0 0 . 0 0 1 3 8 9 − 0 . 0 2 0 8 3 0 . 1 1 8 0 5 6 − 0 . 3 1 2 5 0 . 3 8 0 5 5 6 − 0 . 1 6 6 6 7 0 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 1 2 4 8 1 6 3 2 6 4 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ a b c d e f g ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ 0 . 0 0 1 3 8 9 − 0 . 0 1 2 5 0 . 0 7 6 3 8 9 − 0 . 1 0 4 1 7 0 . 4 2 2 2 2 2 0 . 6 1 6 6 6 7 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ We now have the coefficients to the polynomial: 0 . 0 0 1 3 8 9 x 6 + − 0 . 0 1 2 5 x 5 + 0 . 0 7 6 3 8 9 x 4 + − 0 . 1 0 4 1 7 x 3 + 0 . 4 2 2 2 2 2 x 2 + 0 . 6 1 6 6 6 7 x + 1 Substituting in 13: 0 . 0 0 1 3 8 9 ( 1 3 6 ) − 0 . 0 1 2 5 ( 1 3 5 ) + 0 . 0 7 6 3 8 9 ( 1 3 4 ) − 0 . 1 0 4 1 7 ( 1 3 3 ) + 0 . 4 2 2 2 2 2 ( 1 3 2 ) + 0 . 6 1 6 6 6 7 ( 1 3 ) + 1 = 4 0 9 6 , which is 2 1 3
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@Kevin Luo – the last number is incorrect. 4096=2^12
This is how did it
I used the difference method too with 1s on the 6th difference layer, then constructed each of the other levels by addition. Unfortunately I made arithmetic and reading errors so failed to get 4096. Your analytic metho d is much better, I just didn't know it.
We define: f ( x ) = i = 0 ∑ 6 ( x k ) .
Now, for 0 ≤ x ≤ 6 . we get f ( x ) = 2 x .
Now, f ( 1 3 ) = i = 0 ∑ 6 ( 1 3 i )
On interpreting the summation accordingly, we get:
f ( 1 3 ) = 2 1 i = 0 ∑ 1 3 ( 1 3 i )
Therefore, f ( 1 3 ) = 2 1 2